Ios -[\u NSInlineData objectForKey:]：发送到实例0x7e351cc0'的无法识别的选择器；

我正在尝试从成功登录详细信息中提取用户ID。下面是我的代码`

AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
      manager.requestSerializer = [AFHTTPRequestSerializer serializer];
      manager.responseSerializer=[AFHTTPResponseSerializer serializer];

      [manager.requestSerializer setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];

      manager.responseSerializer.acceptableContentTypes=[NSSet setWithObject:@"text/html"];


      NSString *str=[NSString stringWithFormat:@"{\"username\":\"%@\",\"password\":\"%@\"}",self.usernameTextField.text,self.PswdTextField.text];
      NSDictionary *param=@{@"params":str};


      [manager POST:@"URL" parameters:param

            success:^(AFHTTPRequestOperation *operation, id responseObject) {
                  NSLog(@"responseobject=%@ success=%@",responseObject,operation.responseString);

                NSLog(@"success=%@",[responseObject objectForKey:@"userinfo"]);

我的O/p：

responseobject=<7b22636f 6465223a 302c2275 73657269 6e666f22 3a7b2269 64223a22 31373222 2c226669 7273746e 616d6522 3a224179 75736822 2c226c61 73746e61 6d65223a 22477570 7461222c 22757365 726e616d 65223a22 67757074 61222c22 75736572 656d6169 6c223a22 66686a6a 6473227d 2c226d65 73736167 65223a22 4c6f6769 6e205375 63636573 7366756c 6c79227d> success={"code":0,"userinfo":{"id":"172","firstname":"Ay","lastname":"ta","username":"ss","useremail":"email"},"message":"Login Successfully"}

---

去掉上面写着

manager.responseSerializer=[AFHTTPResponseSerializer serializer];

---

默认情况下，AFNetworking将为您解析JSON。AFHTTPResponseSerializer只会将NSData对象作为responseObject提供给您

manager.responseSerializer=[AFHTTPResponseSerializer serializer];

---

默认情况下，AFNetworking将为您解析JSON。AFHTTPResponseSerializer只会给你一个NSData对象作为responseObject

谢谢你的知识。它可以工作。但是我仍然无法检索用户ID。NSArray*info=[响应对象值forkey:@“userinfo”]；对于（info中的NSString*userid）{self.profile.userid=[userid valueForKey:@“id”]；}仍然发生崩溃[valueforundfinedkey:]：该类不符合密钥id的键值编码。发生什么错误？更正me plzzuserid是字符串（您已在“for”声明中定义了它），但是你试着像字典一样访问它。它可以工作。但是我仍然无法检索用户ID。NSArray*info=[响应对象值forkey:@“userinfo”]；对于（NSString*userid in info）{self.profile.userid=[userid valueForKey:@“id”]；}仍然会发生崩溃[valueforundinedkey:]：此类不符合密钥id的键值编码。“我做错了什么？更正我plzzuserid是一个字符串（您在'for'声明中已将其定义为字符串）。”，但你要像查字典一样去查